Question

Find the smallest number which when increased by $17$ is exactly divisible by $520$ and $468$.

Answer 1

The given numbers are $520$ and $468$
The smallest number which increased by $17$ is exactly divisible by both $520$ and $468$ is obtained by subtracting from the L.C.M of $520$ and $468$.
Prime factorisation of $520=2\times 2\times 2\times 5\times 13$
Prime factorisation of $468=2\times 2\times 3\times 3\times 13$
L.C.M of $520$ and $468$.=$2\times 2\times 2\times 5\times 3\times 3\times 13=4680$
The smallest number which when increased by $17$ is exactly divisible by $520$ and $468$
=$4680-17$
=$4663$